Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\phi\colon[0,1] \to \mathbb R$ be such that $\phi,\phi^\prime,\phi^{\prime\prime}$ are continuous on $[0,1]$, then the following inequality holds:

$$\int_0^1\cos x\frac{x\phi^\prime(x)-\phi(x)+\phi(0)}{x^2}\mathrm dx < \frac32\|\phi^{\prime\prime}\|_\infty$$

I have no idea how to solve this problem, could you help me please?

share|improve this question
    
What does $||\phi''||_{\infty}$ mean?? –  Ramana Venkata Oct 8 '11 at 22:48
    
It's the uniform norm. If $f\colon D\to\mathbb C$, then $\|f\|_\infty:=\sup \{|f(x)| : x\in D\}$. –  SL2 Oct 8 '11 at 22:55
    
Please don't write \mathrm{cos}\;x. Write \cos x (with a backslash) instead. The backslash both prevents italicization and results in proper spacing. With some operators like \max it also causes formatting conventions to be followed, so that in a "displayed" (as opposed to "inline") setting \max_{x\in A} gives you $\max\limits_{x\in A}$. –  Michael Hardy Oct 9 '11 at 2:37

2 Answers 2

up vote 4 down vote accepted

We have \begin{align*} x\phi'(x)-\phi(x)+\phi(0)&=x\phi'(x)-\int_0^x\phi'(t)dt\\ &=\int_0^x\left(\phi'(x)-\phi'(t)dt\right)\\ &=\int_0^x\int_t^x\phi''(s)dsdt, \end{align*} hence $$\left|\frac{x\phi'(x)-\phi(x)+\phi(0)}{x^2}\right|\leq \frac{\lVert \phi''\rVert}{x^2}\int_0^x\int_t^xdsdt=\frac{\lVert \phi''\rVert}{x^2}\int_0^x(x-t)dt=\frac{\lVert \phi''\rVert}2.$$ Therefore, the integral is convergent and
$$\int_0^1\cos x\frac{x\phi'(x)-\phi(x)+\phi(0)}{x^2}dx\leq\frac{\lVert \phi''\rVert}2 \int_0^1\cos xdx =\frac{\lVert \phi''\rVert}2\sin 1\leq \frac{\lVert \phi''\rVert}2,$$ unless I'm misunderstanding something.

share|improve this answer

Use Taylor series twice (I write f for $\phi$ cause I'm lazy):

$f(x) = f(0) + xf'(0) + \frac{x^2}{2} f''(c)$ and $f'(x) = f'(0) + xf''(d)$ where $0 \le c, d \le 1$.

From the second one, $f'(0) = f'(x) - xf''(d)$.

Putting this in the first, $f(x) = f(0) + x(f'(x)-xf''(d)) + \frac{x^2}{2} f''(c)$ so $x f'(x) - f(x) + f(0) = -x^2 f''(d) + \frac{x^2}{2}f''(c)$ or $$\frac{x f'(x) - f(x) + f(0)}{x^2} = -f''(d)+f''(c)/2$$.

Putting this in the integral, since $|\cos| \le 1$ gives the result.

It will be interesting to see how much this agrees with the answer entered while I was entering this

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.